2^-9x=7^x+10
Solve for x and write the exact answer using base-10 logarithms.

Question

2^-9x=7^x+10
Solve for x and write the exact answer using base-10 logarithms.

anonymous · Answer

$$2^{-9x}=7^{x+10}$$

anonymous · Answer

Yeah that's how it is written.

anonymous · Answer

I got some weird funky fraction and I don't think it's right.

anonymous · Answer

Task steps:  Take the log of both sides.  Use the power rule to pull the variables out of the logarithm.  Distribute where possible.  Move terms with x to one side of the equation, other terms to the other side.  Factor out x.  Divide by the other factor to get x=....  It will probably be a mess of logs and numbers, but that's the answer.  If you need an approximation, use your calculator.

anonymous · Answer

What answer did you get?

anonymous · Answer

Does the answer have the term "ln" in it?

anonymous · Answer

10ln(7)/9ln(2)+ln(7)

anonymous · Answer

$$2^{-9x}=7^{x+10} \implies -9x \log 2=(x+10)\log 7 \implies xlog7+9xlog2=-10\log7$$$$\implies x(\log7+9\log2)=-10\log7 \implies x=\frac{-10\log7}{\log7+9\log2}$$

anonymous · Answer

so ln is another term for log, okay thank you!

anonymous · Answer

The specification in the problem said to use common log.  I think your answer is within a negative sign of the correct answer.

anonymous · Answer

ln is the natural log (base e).  The common log is base 10.

anonymous · Answer

Oh okay, thanks for your help!

anonymous · Answer

No sweat.  Do math every day.